2023-5+5+5+5+5+......................................................................................+5
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Cho \(A=\dfrac{2023^{30}+5}{2023^{31}+5}\) và \(B=\dfrac{2023^{31}+5}{2023^{32}+5}\). So sánh A và B
Áp dụng tính chất : Nếu \(\dfrac{a}{b}< 1\) thì \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\) ( a; b; n ϵ N , b; n ≠ 0 )
Ta có \(\dfrac{2023^{31}+5}{2023^{32}+5}< 1\)
⇒ \(B=\dfrac{2023^{31}+5}{2023^{32}+5}< \dfrac{2023^{31}+5+2018}{2023^{32}+5+2018}=\dfrac{2023^{31}+2023}{2023^{32}+2023}=\dfrac{2023\left(2023^{30}+1\right)}{2023\left(2023^{31}+1\right)}=\dfrac{2023^{30}+1}{2023^{31}+1}=A\)Vậy A > B
Ta có 2023A = \(\dfrac{2023.\left(2023^{30}+5\right)}{2023^{31}+5}=\dfrac{2023^{31}+5.2023}{2023^{31}+5}\)
\(=1+\dfrac{2022.5}{2023^{31}+5}\)
Lại có 2023B = \(\dfrac{2023.\left(2023^{31}+5\right)}{2023^{32}+5}=\dfrac{2023^{32}+2023.5}{2023^{32}+5}\)
\(=1+\dfrac{2022.5}{2023^{32}+5}\)
Dễ thấy 202331 + 5 < 202332 + 5
\(\Leftrightarrow\dfrac{2022.5}{2023^{31}+5}>\dfrac{2022.5}{2023^{32}+5}\)
\(\Leftrightarrow1+\dfrac{2022.5}{2023^{31}+5}>1+\dfrac{2022.5}{2023^{32}>5}\)
\(\Leftrightarrow2023A>2023B\Leftrightarrow A>B\)
M=(1/5+1/5^2+1/5^3+...+1/5^2023) + 1/5x(1/5+1/5^2+1/5^3+...+1/5^2022) + ... + 1/5^2021x(1/5+1/5^2) + 1/5^2022x1/5
Xét biểu thức N=1/5+1/5^2+1/5^3 + ... + 1/5^k (K>0, k thuộc Z)
=> 5N=1+1/5+1/5^2+1/5^3+...+1/5^(k-1)
=> 4N= 5N - N =1 - 1/5^k
=> 1/5+1/5^2+1/5^3 + ... + 1/5^k = 1/4x(1-1/5^k)
Thay vào biểu thức M, ta có:
M= 1/4x(1-1/5^2023) + 1/5x1/4x(1-1/5^2022) + ... + 1/5^2021x1/4x(1-1/5^2) + 1/5^2022x1/4x(1-1/5)
=> 4M = (1+1/5+1/5^2+...+1/5^2022) - 2023/5^2023
=> 4M = 5/4x(1-1/5^2023)-2023/5^2023 < 5/4
=> M < 5/16 < 1/3
Vậy M < 1/3 [ vượt chỉ tiêu nhé =)) ]
Lời giải:
$a=1+5+5^2+5^3+...+5^{2022}+5^{2023}$
$5a=5+5^2+5^3+5^4+....+5^{2023}+5^{2024}$
$\Rightarrow 5a-a=5^{2024}-1$
$\Rightarrow 4a=5^{2024}-1$
$\Rightarrow 4a+1=5^{2024}\vdots 5^{2023}$ (đpcm)
\(S=-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\)
\(\Rightarrow\dfrac{25}{5}=-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}\)
\(\Rightarrow5S+S=\left(-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5^2}-...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\right)\)
\(\Rightarrow6S=-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{2021}}-\dfrac{1}{5^{2022}}-\dfrac{1}{5}+\dfrac{1}{5^2}-...+\dfrac{1}{5^{2022}}-\dfrac{1}{5^{2023}}\)
\(\Rightarrow6S=-1-\dfrac{1}{5^{2023}}\)
\(\Rightarrow S=\dfrac{-1-\dfrac{1}{5^{2023}}}{6}\)
A-B
A = 50+52+54+...52022
52xA=52+54+...52024
24xA = 52024-1
A=\(\dfrac{5^{2024}-1}{24}\)
B = 51+53+...52023
B =5x(50+52+...52022) = 5xA
M = A-B = A-5xA = -4A
M=\(\dfrac{1-5^{2024}}{6}\)
Vậy 24xA - 1 = 52024
Nên 52024 chia cho 3 dư 2
Ta có:
\(x^2+5y^2-4x-4xy+6y+5=0\\\Rightarrow[(x^2-4xy+4y^2)-(4x-8y)+4]+(y^2-2y+1)=0\\\Rightarrow[(x-2y)^2-4(x-2y)+4]+(y-1)^2=0\\\Rightarrow(x-2y-2)^2+(y-1)^2=0\)
Ta thấy: \(\left\{{}\begin{matrix}\left(x-2y-2\right)^2\ge0\forall x,y\\\left(y-1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-2y-2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)
Mà: \(\left(x-2y-2\right)^2+\left(y-1\right)^2=0\)
nên: \(\left\{{}\begin{matrix}x-2y-2=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2y+2\\y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot1+2=4\\y=1\end{matrix}\right.\)
Thay \(x=4;y=1\) vào \(P\), ta được:
\(P=\left(4-3\right)^{2023}+\left(1-2\right)^{2023}+\left(4+1-5\right)^{2023}\)
\(=1^{2023}+\left(-1\right)^{2023}+0^{2023}\)
\(=1-1=0\)
Vậy \(P=0\) khi \(x=4;y=1\).
A=5(1+5^2)+5^5(1+5^2)+...+5^2021(1+5^2)
=26(5+5^5+...+5^2021) chia hết cho 26
CÁI 5+5+5+5+5+...................................+5 LÀ 100 SỐ 5
=2023-(5+5+5+5+5........+5)
=2023-(5x100)
=2023-500
=1523